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Let's say I am in Alice's frame.Alice is located on earth. A planet called planet 2 is located 20.3 Light years away. Bob is on a ship that starts on earth and is travelling at a constant velocity .99c. Ignore the acceleration of the ship to get .99c Alice and Bob's ship's clock reads 0 when they reach earth.

Bob travels to the planet 2. Alice observers the time it takes for Bob to reach planet 2 using Alice's clock. What is this elapsed time?

Is the Lorentz transform only used during switching frames? Let me define switching frames. I may have a mistake in my definition. Please correct me if I am wrong. In this example I have 2 frames Alice and Bob. Alice is stationary Bob is moving.

I can calculate Bob's velocity in Alice's frame.

I simply do t= d /v. 20.3 / .99c = 20.5 years. This gives me the elapsed time.

If I wanted to know the time in Alice's frame.

Here is where I get confused.

I have a stationary frame. A moving frame I can switch between frames. When do I switch the stationary frame to become the moving frame, and the stationary frame becomes the moving frame?

When do I stay in the same frame but use the moving frame or the stationary frame?

What is the difference between the two previous questions?

Thanks.

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1 hour ago, can't_think_of_a_name said:

I have a stationary frame. A moving frame I can switch between frames. When do I switch the stationary frame to become the moving frame, and the stationary frame becomes the moving frame?

I'm having difficulty understanding what you mean. Do you understand how you assign coordinates to events in SR?

I don't know what you mean when you say "switching" frames. Jumping from one to another?

That's not how you measure time intervals and distances in SR. It's by sending light rays and receiving them back, and then kind of "triangulating", but taking into account that the signal speed is c no matter what inertial frame you're in.

Let me ask you a question: How do you picture in your mind that Alice knows Bob has already arrived in planet 2?

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51 minutes ago, joigus said:

Let me ask you a question: How do you picture in your mind that Alice knows Bob has already arrived in planet 2?

Surely Alice just reads it off her clock ?

2 hours ago, can't_think_of_a_name said:

I can calculate Bob's velocity  journey time in Alice's frame.

I simply do t= d /v. 20.3 / .99c = 20.5 years. This gives me the elapsed time.

If I wanted to know the time in Alice's frame.

Here is where I get confused.

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3 hours ago, can't_think_of_a_name said:

Let's say I am in Alice's frame.Alice is located on earth. A planet called planet 2 is located 20.3 Light years away. Bob is on a ship that starts on earth and is travelling at a constant velocity .99c. Ignore the acceleration of the ship to get .99c Alice and Bob's ship's clock reads 0 when they reach earth.

Bob travels to the planet 2. Alice observers the time it takes for Bob to reach planet 2 using Alice's clock. What is this elapsed time?

Is the Lorentz transform only used during switching frames? Let me define switching frames. I may have a mistake in my definition. Please correct me if I am wrong. In this example I have 2 frames Alice and Bob. Alice is stationary Bob is moving.

I can calculate Bob's velocity in Alice's frame.

I simply do t= d /v. 20.3 / .99c = 20.5 years. This gives me the elapsed time.

If I wanted to know the time in Alice's frame.

Here is where I get confused.

I have a stationary frame. A moving frame I can switch between frames. When do I switch the stationary frame to become the moving frame, and the stationary frame becomes the moving frame?

When do I stay in the same frame but use the moving frame or the stationary frame?

What is the difference between the two previous questions?

Thanks.

All calculations of distance and time have to be done in the same frame.

You use the transforms to get the values in the other frame. You have done the calculations in Alice’s frame. If you want the distance and/or time in Bob’s frame, you do the transform.

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19 minutes ago, studiot said:

Surely Alice just reads it off her clock ?

Exactly! I want the OP to think: How do I measure separation of distant things? By measuring times.

Observers cannot read space directly. How can distant "news reporters" at rest with respect to us liaise with us?  It takes some time to realise that this is necessary.

I hope this sketch clarifies the meaning of coordinates in SR: $t=\frac{1}{2}\left(t_{1}+t_{2}\right)$

$x=\frac{c}{2}\left(t_{2}-t_{1}\right)$

So everything, spacial coordinates included, is based on observation times.

I'm not sure if that's what confusing you, @can't_think_of_a_name. It looks to me like that may be it.

Once it's clear that you've got all of space-time mapped with these x's and t's, you do what Swansont is telling you and apply Lorentz transformations. But you don't get information back from distant events by sending someone back that "switches" inertial frames. It's a matter of sending light rays back and forth what gives meaning to the coordinates, not people on a spaceship with the news.

I hope that helped and had to do with the source of your confusion.

1 hour ago, studiot said:

4 hours ago, can't_think_of_a_name said:

I can calculate Bob's velocity  journey time in Alice's frame.

I simply do t= d /v. 20.3 / .99c = 20.5 years. This gives me the elapsed time.

If I wanted to know the time in Alice's frame.

Here is where I get confused.

+1. Thank you. That got me confused too.

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28 minutes ago, joigus said:

Exactly! I want the OP to think: How do I measure separation of distant things? By measuring times.

This is not how I understand the original situation.

I understand it to mean that

The distance from Earth to planet 2 has already been measured (and will not change in the lifetime of Alice) by othe astronomical means.

Similarly the spaceship has a 'cruise control so its velocity is predetermined at a fixed rate of 0.99c. No one measures this.

Bob starts out from Earth when both his clock and Alice's clock read zero.

So by Alice will know that when her clock reads 20.5 years Bob is just arriving at planet 2 with his clock.

Bob's clock will be reading a value given by the Lorenz transform of this time.

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30 minutes ago, studiot said:

This is not how I understand the original situation.

I understand it to mean that

[...]

I see. And what about the switching frames? Did you understand that part?

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1 hour ago, studiot said:

Similarly the spaceship has a 'cruise control so its velocity is predetermined at a fixed rate of 0.99c. No one measures this.

Or it doesn’t matter, since it’s a relative speed, and both Alice and Bob will agree on it.

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Posted (edited)

Let me clarify because my first situation was not explained well.

Please refer to the picture from imgur.

Find the time of Alice's clock when B reaches planet 2?
Find the time of Bob's clock when B reaches planet 2?

First question I think I understand . I just use d= vt  t = d/v
Second question.
Shouldn't the time of B = A? But I know this is wrong because moving clocks tick slow. Why do I have to perform the Lorentz transform when from my picture it shows it as the same time but I know the Lorentz transform must be performed because Alice is moving.

I think the answer is because I can only measure things in stationary frames and the Lorentz transform does this transform but also twists the graph . Since the graph is twisted during a Lorentz transform even if it is on the T_B axis  T_B = 0  isn't 0. Is this correct?

Edited by can't_think_of_a_name

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8 hours ago, joigus said:

I see. And what about the switching frames? Did you understand that part?

Yes, but I didn't want to introduce the second part of the development before the first was digested.

This comment also applies to swansont's more complete and comprehensive statements.

10 hours ago, swansont said:

All calculations of distance and time have to be done in the same frame.

You use the transforms to get the values in the other frame. You have done the calculations in Alice’s frame. If you want the distance and/or time in Bob’s frame, you do the transform.

8 hours ago, swansont said:

Or it doesn’t matter, since it’s a relative speed, and both Alice and Bob will agree on it.

Yes the next part of my offering would be for Bob to calculate his version of the journey and apply Lorenz to find his measurement of the distance travelled and thus find the same relative velocity as Alice observed.

Then one could enquire about how the spaceship achieves constant velocity.

5 hours ago, can't_think_of_a_name said:

Let me clarify because my first situation was not explained well.

Please refer to the picture from imgur.

Find the time of Alice's clock when B reaches planet 2?
Find the time of Bob's clock when B reaches planet 2?

First question I think I understand . I just use d= vt  t = d/v
Second question.
Shouldn't the time of B = A? But I know this is wrong because moving clocks tick slow. Why do I have to perform the Lorentz transform when from my picture it shows it as the same time but I know the Lorentz transform must be performed because Alice is moving.

I think the answer is because I can only measure things in stationary frames and the Lorentz transform does this transform but also twists the graph . Since the graph is twisted during a Lorentz transform even if it is on the T_B axis  T_B = 0  isn't 0. Is this correct?

Did you follow the first steps of my explanation, which give the correct sequence to consider things ?

Quote

Find the time of Alice's clock when B reaches planet 2?
First question I think I understand . I just use d= vt  t = d/v

We have all said this is correct.

Quote

Find the time of Bob's clock when B reaches planet 2?
Second question.
Shouldn't the time of B = A?

No this is not correct and is the subject of the second part of my explanation that I have not yet provided.

You need to fully understand why you have got the first part right before moving on to the second.

Hopefully you will also take the opportunity to learn some correct terminology as you do this.

'Twist' is the wrong word. You should use 'Rotate'.
If you swing a weight on a rope in a circle, the rope is straight not twisted.
But it is rotating about an axis that is not along the axis of the rope.

A twist would be about the axis of the rope.

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Posted (edited)
11 hours ago, joigus said:

$t=\frac{1}{2}\left(t_{1}+t_{2}\right)$

$x=\frac{c}{2}\left(t_{2}-t_{1}\right)$

So everything, spacial coordinates included, is based on observation times.

On further reflection, after reading Studiot's comments (+1) carefully, and to correct myself: This definition, while it's very widely used theoretically and very solid, is not universal; and in particular is not useful to assign space coordinates to distant planets. Also, it had nothing to do with the OP's confusion.

Sorry, it was late, my reasoning was obfuscated by the phrasing of the question, and consequently I wasn't very helpful.

Edited by joigus
minor correction (semantics)

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7 hours ago, can't_think_of_a_name said:

First question I think I understand . I just use d= vt  t = d/v
Second question.
Shouldn't the time of B = A? But I know this is wrong because moving clocks tick slow. Why do I have to perform the Lorentz transform when from my picture it shows it as the same time but I know the Lorentz transform must be performed because Alice is moving.

I think the answer is because I can only measure things in stationary frames and the Lorentz transform does this transform but also twists the graph . Since the graph is twisted during a Lorentz transform even if it is on the T_B axis  T_B = 0  isn't 0. Is this correct?

If you want Bob’s numbers, you either analyze the experiment in his frame, or transform from Alice’s frame.

One thing we know is the answers will be different in each frame. In Bob’s frame, the trip will be about 3LY and consequently take about 3 years.

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Posted (edited)
10 hours ago, swansont said:

Or it doesn’t matter, since it’s a relative speed, and both Alice and Bob will agree on it.

True. +1

Edited by joigus

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10 hours ago, swansont said:

If you want Bob’s numbers, you either analyze the experiment in his frame, or transform from Alice’s frame.

One thing we know is the answers will be different in each frame. In Bob’s frame, the trip will be about 3LY and consequently take about 3 years.

I think your comment and my comment state Bob always has to  be stationary. Is this correct?  Another member mentioned I should use rotate instead of twist. So since I rotate the graph XB = 0.  X_B is different to 0 compared to the non rotated graph giving a value different then 0. Is this correct? Or are you stating something different?

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4 minutes ago, can't_think_of_a_name said:

I think your comment and my comment state Bob always has to  be stationary. Is this correct?  Another member mentioned I should use rotate instead of twist. So since I rotate the graph XB = 0.  X_B is different to 0 compared to the non rotated graph giving a value different then 0. Is this correct? Or are you stating something different?

Any observer, Bob included, is stationary in their own frame.

8 minutes ago, can't_think_of_a_name said:

Another member mentioned I should use rotate instead of twist. So since I rotate the graph XB = 0.  X_B is different to 0 compared to the non rotated graph giving a value different then 0. Is this correct? Or are you stating something different?

12 hours ago, studiot said:

Did you follow the first steps of my explanation, which give the correct sequence to consider things ?

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8 minutes ago, can't_think_of_a_name said:

I think your comment and my comment state Bob always has to  be stationary. Is this correct?  Another member mentioned I should use rotate instead of twist. So since I rotate the graph XB = 0.  X_B is different to 0 compared to the non rotated graph giving a value different then 0. Is this correct? Or are you stating something different?

Bob is stationary in his frame. You always need to realize what frame you are in when analyzing.

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12 hours ago, joigus said:

Not a very useful line of thinking for homework help.

I agree there is a great deal of off topic discussion which makes it difficult for someone studying relativity from the beginning and for those trying to help that person (the OP).

+1

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Just now, studiot said:

I agree there is a great deal of off topic discussion which makes it difficult for someone studying relativity from the beginning and for those trying to help that person (the OP).

+1

Thanks. And you're dead right. +1. I stepped back on this particular topic because Swansont and you were more focused and had this more under control and I wasn't helping. I was "waxing theoretical".

I think that's the idea.

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I think I get it now.

I am just going to the do the math. t = d/v. t = 20.5  if memory is correct. Since Bob is moving I perform the Lorentz transform for time.  T = gamma(T' - (v/c^2)(x)

Gamma is = 7.14 if memory serves correct.

20.5 year  = 7.14 (T' - .99c/c^2)(20.3) =

T = 7.14 (T' -  20.1) =

T = 7.14( -20.1T' )

I think I made a mistake T' = gamma(T - (v/c^2)(x).

The mistake x = 0.

T = 7.14 (20.5 years) - .(99c/c^2)(0) =

T = gamma (T') =

T' = T /  gamma =

T' = 20.5 / 7.14 =  2.87

Is this correct?

Any further questions I will ask.

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